= Definite integral
{title2=$\int_a^b f(x)\,dx$}
A <definite integral> accumulates an <integrand> between specified endpoints, retaining the sign of each contribution. For a <real-valued function> on a finite <real interval>, a finite <definite integral> is a <scalar>; an <indefinite integral> describes a family of <antiderivatives>. The lower and upper limits of integration $a,b$ specify where accumulation starts and ends. The integration variable is a placeholder: $\int_a^b f(x)\,dx=\int_a^b f(u)\,du$.
This is useful whenever a varying rate must be converted into a total change. For motion along a <straight line> with <continuous> <velocity> $v(t)$ and <position> $s(t)$ satisfying $s'(t)=v(t)$, the <fundamental theorem of calculus> gives <displacement>, while accumulating <speed> gives the <distance traveled from velocity>:
$$
s(b)-s(a)=\int_a^b v(t)\,dt,\qquad
D=\int_a^b |v(t)|\,dt,\qquad a<b.
$$
Negative <velocity> reduces <displacement> while adding to the <distance traveled from velocity>. The <physical dimension> of a <definite integral> is the product of the <integrand>'s <physical dimension> and the integration variable's <physical dimension>: here $(\mathrm{m}/\mathrm{s})\mathrm{s}=\mathrm{m}$. An <area> on a rate graph need not be a spatial <area>.<Calculus Volume 1 by Gilbert Strang and others/5.4 Integration Formulas and the Net Change Theorem>[Theorem 5.6 and the displacement and total-distance discussion]{ref}
As an explicitly synthetic motion model, let $v(t)=(2\,\mathrm{m}\,\mathrm{s}^{-2})t-2\,\mathrm{m}\,\mathrm{s}^{-1}$ for $0\le t\le3\,\mathrm{s}$, with $s(0)=0$. Using numerical values in seconds and metres, an <antiderivative> gives $s(t)=t^2-2t$. The <velocity> changes sign at $t=1$: the model first moves $1$ m in the negative direction, then $4$ m in the positive direction. Its <displacement> is $-1+4=3$ m, and its <distance traveled from velocity> is $1+4=5$ m. The initial <position> fixes the <constant of integration>, but does not change either accumulated quantity.
\Image[/variable-velocity-displacement-and-distance.png]
{title=Synthetic variable velocity model with displacement 3 metres and distance traveled 5 metres over three seconds}
{id=variable-velocity-displacement-and-distance}
Orientation matters independently of whether the <integrand> is positive or negative. For a <function> <integrable> on the <interval> containing the endpoints, use the conventions
$$
\int_a^a f(x)\,dx=0,\qquad
\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx,\qquad
\int_a^c f(x)\,dx=\int_a^b f(x)\,dx+\int_b^c f(x)\,dx.
$$
For the preceding <velocity> model, accumulation from $3$ seconds back to $0$ seconds gives $-3$ m. Reversing the bounds of the <speed> <integral> gives $-5$ m, which is an oriented <integral>, not a negative <distance traveled from velocity>; physical elapsed-time totals use the earlier endpoint first. The splitting rule holds even when $b$ is outside the <interval> between $a$ and $c$, provided <integrability> holds on the <interval> containing all three endpoints.<Calculus Volume 1 by Gilbert Strang and others/5.2 The Definite Integral>{ref}
Existence needs hypotheses. For a <bounded function> $f$ on $[a,b]$ with $a<b$, a finite <Riemann integral> exists when every tagged <Riemann sum> approaches the same value as the largest cell width tends to zero:
$$
\int_a^b f(x)\,dx=\lim_{\max_i(x_i-x_{i-1})\to0}
\sum_{i=1}^n f(\xi_i)(x_i-x_{i-1}),\qquad
a=x_0<\cdots<x_n=b,\quad \xi_i\in[x_{i-1},x_i].
$$
The <limit> must be independent of both the <partition of an interval> and the tags. Merely making the number of cells large can leave a wide cell unresolved. On a finite <closed interval>, every <continuous function>, and every <bounded function> that is <continuous> except at finitely many points, is <Riemann integrable>.<Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl/5.2 Properties of the integral>[Lemma 5.2.7 and Theorem 5.2.9]{ref} Boundedness alone is insufficient: the <Dirichlet function> on $[0,1]$ is $1$ on the <rational numbers> and $0$ on the <irrational numbers>. Tags can all be chosen rational or all irrational, giving <Riemann sums> of $1$ or $0$ at arbitrarily small cell widths.<Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl/5.1 The Riemann integral>[Example 5.1.4]{ref}
For a finite signed <Lebesgue integral>, the condition is instead that $f$ be a <measurable function> for <Lebesgue measure> and $\int_{[a,b]}|f|\,dx<\infty$. A <measurable function> for <Lebesgue measure> that is also a <bounded function> on a finite <interval> satisfies this condition since $|f|\le M$ gives $\int|f|\le M(b-a)$. The <Dirichlet function> therefore has <Lebesgue integral> zero: the <rational numbers> form a <null set>, so it equals zero <Lebesgue almost everywhere>. The <Riemann integral> and <Lebesgue integral> agree for the <continuous functions> used in the diagrams.<18.102 Introduction to Functional Analysis, Spring 2021 (lecture notes) by Andrew Lin/Lecture 12: Lebesgue Integrable Functions, the Lebesgue Integral and the Dominated Convergence Theorem>[Definition 120, Proposition 123(2), Remark 124 and Theorem 126]{ref} An unbounded <integrand> or infinite endpoint requires further analysis, for example the <limits> defining an <improper integral>; the notation alone does not establish convergence.
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